How to use automorphisms to produce isotrivial non trivial families

Question

In many places the existence of automorphism is acknowledged as one of the reasons why fine moduli spaces cannot exist. A typical example is the following.

Consider a curve $C$ with a nontrivial automorphism, for instance a hyperelliptic curve with its involution $\phi$. Now let $B$ be any scheme with a free action of (in this case) $\mathbb{Z}/(2)$.

Let $D$ be the quotient of $B \times C$ by the diagonal action of $\mathbb{Z}/(2)$ and let $B'$ be the quotient of $B$ by the action of $\mathbb{Z}/(2)$.

Consider the map $f \colon D \rightarrow B'$. Then the fibers of $f$ are all isomorphic to $C$, but for a suitable choice of $B$ the family $f$ is not isomorphic to the product $C \times B'$.

How can I get an explicit example of the last assertion?

How can I produce $B$ as above such that $f$ is not isomorphic to the trivial family?

Answer 1

Here is an explicit example. Denote by $E$ the elliptic curve $E=\mathbb C/(\mathbb Z+i\mathbb Z)$. Now let $B=C=E$. The involution that we will consider is $$(z,w)\in (E\times E)\to (z+\frac{1}{2},-w).$$

Notice now that this quotient is not a direct product anymore. You can see this by calculating $H_1(E\times E)/\mathbb Z_2$ and verifying that it is not $\mathbb Z^4$. Alternatively, you can notice that the quotient does not have a non-vanishing holomorphic volume form. Indeed if such form existed it could be lifted to $E\times E$ to a form $cdz\wedge dw$ ($c\ne 0$), which would give a contradiction since $dz\wedge dw$ is anti-invariant.

Answer 2

If $C=B$ with the same $Z/2$ action, and $g(C/Z_2) \geq 2$, then you always get a non-trivial family:

Suppose it was trivial. Then it had another projection $q : D \to C$. Since $C$ is an \'etale cover of $B'$, and both have genus at least two, $g(C) > g(B')$ by the Hurwitz-formula. That is, also by the Hurwitz formula, all the maps from $B'$ to $C$ are the constant maps. So, all sections $B' \to D$ of $f$ would have to be contained in a fiber of $q$, i. e. they have to be one fiber of $q$. However there are two natural sections $E$ and $F$ of $p$: $[c] \to [(c,c)]$ and $[c] \to [(c,-c)]$. So both of these have to be contracted to a point by $q$. Given any $b \in B$, a choice $c \in C$ such that $[c]=b$ gives an isomorphism $a_c : p^{-1}(d) \to C$, which sends $[(c,c')]$ to $c'$. Now one can construct automorphisms $\varphi_{c'}$ of $C$ for any $c' \in C$ by the following composition:

$ C \to^{a_c^{-1}} p^{-1}([c]) \to^q C \to^s p^{-1}([c']) \to^{a_{c'}} C $

where

$ S=(q|_{p^{-1}([c'])})^{-1} $

One specialty about $\varphi_{c'}$ is that it takes $c$ and $-c$ to $c'$ and $-c'$, respectively. This follows from the fact, that $E$ and $F$ are contracted to a point by $q$. So, $C$ has infinitely many different automorphisms, which is a contradiction by the assumption that $g(C) \geq 2$. That is our assumption is false, therefore $Y$ is not a product family.